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2011 Exam
You can find the 2011 Exam here Multiple choice answers are given. Section II, III, IV, V, VI can be discussed below. Please respect the copyright of The University of Melbourne. Section I 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Question 1 (6 marks total) Looks like Guanine (G) on the left and Cytosine © on the right Hydrogen bonds form between the oxygen the hydrogen up top, the hydrogen and the nitrogen in the middle, and the hydrogen and the oxygen down below as per this picture http://www.biosci.ohiou.edu/introbioslab/Bios170/170_8/gc.GIF Recognition sites A = HA, B=HA, C=HD, D=H E=HA, F=HD, G=HA Question 2 (5 marks total) a) 1. Nucleosome 2. Nucleofilaments/ 30nm fibre 3. Chromosome (alternate answer: accordding to lehninger4th, p943, it is a 'rosette') b) 1. False, missing the H4 subunit and there is no H3A and H3B, only H3. 2. True, this is what allows eukaryotes to replicate DNA without gyrase - the nucleosome underwinds the DNA. No, this is FALSE. There are only about 200 base pairs around a eukaryotic nucleosome, which is the approximately the same length as a eukaryotic Okazaki fragment. E. Coli Okazaki fragments are about 1000 - 2000 base pairs long. 3. True, it is double stranded 4. False, the DNA is underwound in histones. The ADJACENT section is overwound, but relaxed by topoisomerases. 5. True, H1 is needed in the DNA between histones - the linker DNA. 6. True, but confusing. CAF-1 IS involved in NUCLEOSOME assembly and a nucleofilament could not form without it. In this sense, it definitely "helps" False- Look at question 3 on the multiple choice. 7. True, in any other phase chromosomes would not allow gene expression because they do not allow the DNA to be opened up in a transcription bubble. Question 3 (9 marks total) OriC: "Origin of Replication". 245bp long. ii) 1. True. These proteins overwind the DNA at A to promote underwinding at B. 2. True? False - FIS and IHF are also in Region A. 3. True. DUE stands for DNA unwinding element and is rich in AT because 2 hydrogen bonds are easier to melt than three. 4. True 5. False. I think this is false, as it can hydrolyse ATP-ADP to which is how I sites release, but i dont think it can convert ADP back to ATP, there isnt any reason for it to. I agree. The DnaA-ATP to DnaA-ADP causes the DnaA to drop off the I1-3 portions, and the replication bubble can close back up more easily. 6. True. iii) In the image in the lectures there are two arrows pointing in a downwards direction on both sides of the "A" in the image above. We woulde expect gyrase to work AHEAD of the replication fork, and since there are two forks, I would argue that gyrase would also act in the lower left corner of the image. -MW iv) GATC (eleven repeats) v) N^6 methyladenine vi) DAM - DNA adenine methylase vii) False. It can only begin when the A in GATC is METHYLATED. Question 4 (6 marks total) (a) Top - Coding Strand Opposite strand- Template strand Segment attached to ehe template strand- mRNA 17 bp bubble 5' RNA 8bp hybrid Bubble is moving right Next one is added 3' on the RNA strand. (b) 1. false - sigma-70 not sigma-54 "brutal question - sigma-54 is the Nitrogen limiting promoter, sigma-70 is the housekeeping promoter http://en.wikipedia.org/wiki/Sigma_factor" - MW 2. false - its bound/free 3. true 4. true Question 5 (a+b) (9 marks total) a) The best way to know the answer is to look at the Lac Operon lecture but here is the general order of elements O3, Promoter begin, CRP (activator), Promoter end, O1, LacZ, O2, LacY, LacA Coding segment begins at LacZ, and is polycistronic. This indicates that ribosomes are able to initiate protein synthesis by binding to any one of three sites on the Lac Operon mRNA strand. I do not think the protein is cleaved after synthesis. *There are S-D sequences present in front of all three separate lac genes, which enable the formation of polysomes (multiple ribosomes binding onto the transcript). There are polycistronic RNA transcripts (e.g: viral RNA transcript of poxvirus) that translate a precursor protein which would then be cleaved into separate functional proteins, but lac operon-derived transcript doesn't work this way. -SA b) FALSE. The "recognition helix" on the Lac repressor binds to ONE HALF of the operator palindrome. True False, you need four. False (couldnt this be true as it is a weak promotor and often leakage occurs?) No, this cannot be true. If repressor is bound to the operator, there is NO expression (even if there are high levels of cAMP). cAMP binds to CRP protein and this complex binds to the CRP site to stabilize RNAP binding at the promoter and increase gene expression IF the respressor isn't bound. There is one exception- Sometimes, the repressor may spontaneously detach from the operator without allolcatose, allowing some expression of the gene (leaky expression), which is why the cell is ready to import and metabolize lactose when required. However, high levels of cAMP cannot overcome repression. Only the presence of allolactose can. Question 6 (10 marks total) incorrect one is C (a) earlier stage: B (initiation) IF-1: prevent tRNA binding to A site IF-2~GTP: guides tRNA^fmet to P site IF-3: prevents 50S binding Energy source: GTP Shine-Dalgarno sequence: aligns AUG in P site (b) later stage: A (elongation) Energy source: GTP NOTE: I'm not sure if GTP is correct . . . It says on the ppt slides (lect 29, slide 9) that the energy source is provided by the high energy link between the aa and its tRNA aa~tRNA EF-Tu~GTP: helps tRNA enter A site EF-TS: recycles EF-TU~GTP translocase (EF-G): moves P-->A site + mimics tRNA~EF-Tu ( I don't think this is necessary as in elongation not translocation step) peptidyl transferase (ribozyme, part of 23S) : makes peptide bond between a.a (Carbonyl end of A site with N terminus of P site amino acid. aminoacyl-tRNA synthetase: x20 per a.a/tRNA. binds to codon in A site SECTION III (25 minutes) a) V, F, A... b) I think hydrogen bonds only occur between the NH of one residue and the C=O of the opposite strand's residue. Refer to this diagram http://www.chembio.uoguelph.ca/educmat/phy456/gif/sheet1.gif c) Beta sheets require all trans bonds, so omega= -180 d) From the lecture notes, phi=-130, psi=130. These angles can be seen on the Ramachandran plot e) Hydrophobic interactions, Van der Waals. You would expect to find them in the core because of the high hydrophobic chacteristics of the side chains. f) Beta-alpha-Beta hairpins. Beta-Barrels is incorrect because they are ANTIPARALLEL in a beta-barrel. SECTION IV (15 minutes) Table 1: 4 Steady state when the production of ES = to removal so KfES= KbES + Kcat ES simplify to get 4 Table 2: In response to "3? because is uses micromol in the information? or 1 so that it is consistent with answer 1 in the 2009 exam...?" I think the turnover number is by default referring to (moles/time of product) / (moles of enzyme). So 1 is correct. -->Is it option 5 because the rate is 800/sec for the enzyme; therefore, one enzyme molecule turns over 800 molecules of product per second = 48,000 per minute??? Yeah i agree thought it was 5, as the definition of a turnover number is per catalytic site of enzyme so would only be one molecule of enzyme. So then 2009 answer would be 4. ---> T/O number: is the max. number of molecules of substrate that an enzyme converts to product.....doesnt help with choosing an answer though ha It's 5. Both these statements imply each other. "1 mole '''of fumarase produces 48000 '''moles of product per minute" "1 '''molecule '''of fumarase produces 48000 '''molecules '''of product per minute" (5) 5 is the only one option that doesn't conflate moles and molecules. Table 3: 3. Vmax=2000. 2000/2=1000. look across to substrate that refers to 1000. Km = 0.5 Table 4: 1. Vmax/Km. 2000/0.5 = 4000 I agree with the answer. This part is a little strange. The specificity constant = Kcat/Km. Vmax=Kcat*Et since this graph has a vertical axis of V/Et (you can check the units), the Vmax on the graph will actually equal Kcat. therefore specificity constant = Kcat/Km= 2000/0.5=4000 But Kcat = 800sec-1or 4800min-1 as per Table 2??? SECTION V (15 minutes) Question 1 (3 marks) polar group--phophate--glycerol-- two fatty acid chains ( not sure if they want the structures of glcerol and phophate or if they want us to state a particular polar group. It said general so I took that to mean we didnt have to be too specific) Question 2 (4 marks) slide 34 from Lecture 18 Question 3 (4 marks total) a) the hydropathy index refers to the level of hydrophobicity of an amino acid residue b) this protein is likely to be an integral membrane protein since it has many transmembrane domains indicated by the stretches of 20 or more hydrophobic residues c) the amino acid residues in the shaded region are hydrophobic they are characterised by: - alliphatic side chains such as valine - aromatic side chains with no ionisable substituents such as phenylalanine - non-polar - non ionic d) Protein Y is likely to be a membrane receptor since it has 7 transmembrane domains ( indicated by the 7 peaks of the graph). Many membrane receptors are 7 transmembrane domain proteins I think it as actually a MEMBRANE TRANSPORTER. The figure corresponds to the Hydropathy Plot for Bacteriorhodopsin from slide 27 lecture 18. Bacteriorhodopsin is a light driven H+ pump. Question 4 (4 marks total) a) linkage is beta 1-->2 ( above the plane of the ring on the non reducing end and bound to the oxygen on the second position) b) same as the galactose ( molecule 1) except that the hydroxy on the C1 position points downwards since it is alpha. also the C2 hydroxy points up since it is a C2 epimer of galactose c) everything but the GalNAc SECTION VI (20 minutes) Question 1. (5 marks) 1. ATP - adenosine triphosphate 2. NAD - nicotinamide adenine dinucleotide 3. glycogen 4. pyruvate 5. UDP-glucose 6. glucose-6-phosphate 7. fructose-1,6-bisphosphate 8. 3-phosphoglycerate 9. Acetyl-CoA 10. citrate Question2. (5 marks) box 1: 2 molecules of ATP are consumed box 2: I think 0 molecules of ATP are produced.- For this question, I think it is 4ATP produced as 2ATP are consumed in glycolysis, 4ATP are made in GNG as there are now 2 molecules of G-3-P (due to Triose Phosphate isomerase). this image suggests that you can get from glyceraldehyde-3-phosphate to glucose without involving any ATP. pathway is called GLUCONEOGENESIS Enzymes 1-7: 1. hexokinase 2. glucose-6-phosphatase 3. phosphohexose isomerase 4. phosphofructokinase-1 5. fructose-1,6-bisphosphatase 6. aldolase 7. triosephosphate isomerase Question 3. (3 marks) Write short answers for the following: i) the total entropy (disorder) of the universe always increases ii) -30.5kJ/mol iii) acetoacetate, beta-hydroxybutyrate, acetone Question 4. (2 marks) 1. NADH 2. CO2 3. FADH2 4. fumarate Question 5. (5 marks) (look at lecture slide 116) **** Makes sure you learn 117 slide though, thats the one which will more than likely be on the exam